An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Analyze the solubility of reactants and products in each reaction using the given solubility rules.
Identify reactions where soluble reactants form at least one insoluble product (precipitate).
The precipitation reactions are: N a 2 S + F e B r 2 − 2 N a B r + F e S , M g S O 4 + C a C l 2 − M g C l 2 + C a S O 4 , and A g N O 3 + N a Cl − A g Cl + N a N O 3 .
Therefore, the final answer is: N a 2 S + F e B r 2 − 2 N a B r + F e S , M g S O 4 + C a C l 2 − M g C l 2 + C a S O 4 , A g N O 3 + N a Cl − A g Cl + N a N O 3 .
Explanation
Understanding the Problem We are given a set of solubility rules and five chemical equations. Our task is to identify which of these equations represent precipitation reactions. A precipitation reaction occurs when two soluble reactants combine to form at least one insoluble product. We will use the solubility rules to determine the solubility of each compound in the reactions.
Analyzing Reaction 1 Let's analyze each reaction:
Reaction 1: N a 2 S + F e B r 2 → 2 N a B r + F e S
N a 2 S : Soluble (Rule 1 - contains N a + )
F e B r 2 : Soluble (Rule 3 - most bromides are soluble)
N a B r : Soluble (Rule 1 - contains N a + )
F e S : Insoluble (Sulfides are generally insoluble, except with Group 1 and ammonium)
Since the reactants are soluble and one of the products ( F e S ) is insoluble, this is a precipitation reaction.
Analyzing Reaction 2 Reaction 2: M g S O 4 + C a C l 2 → M g C l 2 + C a S O 4
M g S O 4 : Soluble (Rule 4 - most sulfates are soluble)
C a C l 2 : Soluble (Rule 3 - most chlorides are soluble)
M g C l 2 : Soluble (Rule 3 - most chlorides are soluble)
C a S O 4 : Insoluble (Rule 4 - sulfates are insoluble with C a 2 + )
Since the reactants are soluble and one of the products ( C a S O 4 ) is insoluble, this is a precipitation reaction.
Analyzing Reaction 3 Reaction 3: L i O H + N H 4 I → L i I + N H 4 O H
L i O H : Soluble (Rule 1 - contains L i + )
N H 4 I : Soluble (Rule 1 - contains N H 4 + )
L i I : Soluble (Rule 1 - contains L i + )
N H 4 O H : Soluble (Rule 1 - contains N H 4 + )
Since all reactants and products are soluble, this is not a precipitation reaction.
Analyzing Reaction 4 Reaction 4: 2 N a Cl + K 2 S → N a 2 S + 2 K Cl
N a Cl : Soluble (Rule 1 - contains N a + )
K 2 S : Soluble (Rule 1 - contains K + )
N a 2 S : Soluble (Rule 1 - contains N a + )
K Cl : Soluble (Rule 1 - contains K + )
Since all reactants and products are soluble, this is not a precipitation reaction.
Analyzing Reaction 5 Reaction 5: A g N O 3 + N a Cl → A g Cl + N a N O 3
A g N O 3 : Soluble (Rule 2 - contains N O 3 − )
N a Cl : Soluble (Rule 1 - contains N a + )
A g Cl : Insoluble (Rule 3 - chlorides are insoluble with A g + )
N a N O 3 : Soluble (Rule 2 - contains N O 3 − )
Since the reactants are soluble and one of the products ( A g Cl ) is insoluble, this is a precipitation reaction.
Identifying Precipitation Reactions Based on our analysis, the precipitation reactions are:
N a 2 S + F e B r 2 → 2 N a B r + F e S
M g S O 4 + C a C l 2 → M g C l 2 + C a S O 4
A g N O 3 + N a Cl → A g Cl + N a N O 3
Final Answer Therefore, the equations that represent precipitation reactions are:
N a 2 S + F e B r 2 − 2 N a B r + F e S
M g S O 4 + C a C l 2 − M g C l 2 + C a S O 4
A g N O 3 + N a Cl − A g Cl + N a N O 3
Examples
Understanding precipitation reactions is crucial in various fields. For instance, in water treatment, precipitation is used to remove impurities by adding chemicals that form insoluble compounds with the contaminants, which can then be filtered out. Similarly, in chemical analysis, precipitation reactions are used to identify the presence of specific ions in a solution. In everyday life, the formation of scale in kettles and pipes is an example of unwanted precipitation, where calcium and magnesium carbonates precipitate out of hard water when heated.
